Let I be the incenter of ABC. AI, BI, CI intersect the circumcircle of ABC again at A', B', C' respectively. Show that the area of A'B'C' >= the area of ABC

In my earlier comment I noted that I had made a rough drawing.  It had too problems, one it was rough and two, my first triangle had acute angles.   This gave a distinct impression that the "greater than" part of the expression was invalid.

I drew this up (still some minor inaccuracies but decent enough to see the reality) using CAD program.  I chose as my triangle an obtuse scalene where the shortest side was about 1/3 of the next longer.

In my diagram the circumcircle is outside of my original scalene but lies within the triangle A'B'C'.  It seems clear to me that if the incentre and circumcentre of a triangle are at the same point then the "projected" (by the rules stated) triangle will be congruent, but as these two centres are allowed to diverge then the "projected" triangle will grow in area.

Unfortunately the visual is not satisfactory proof!

Posted by brianjn on 2007-11-08 07:30:50